Question

Find the tension (in $N$) acting on $2m$ mass when the system is released from rest. Assume that the pulleys are ideal and strings are inextensible.

• A. $\frac{24mg}{29}$
• B. $\frac{23mg}{24}$
• C. $\frac{25mg}{23}$
• D. $\frac{23mg}{25}$

Answer 1

The correct option is A $\frac{24mg}{29}$

The given system can be reframed as shown

From constrained condition and using method of intercept we get,

a) ${\stackrel{..}{L}}_1+{\stackrel{..}{L}}_2=0\Rightarrow -a_1+a_p=0$
$\Rightarrow a_1=a_p$

b) ${\stackrel{..}{L}}_3+{\stackrel{..}{L}}_4=0$
$\Rightarrow -a_2-a_p-a_p+a_3=0$
$\Rightarrow 2a_p=a_3-a_2$
$\Rightarrow 2a_1=a_3-a_2$.............(1)

From FBD of $2m,3m$ and $m$
$3m:3mg-T_2=3m{a}_3\dots \dots \left(2\right)$ and $T_1=2T_2\dots \dots \left(3\right)$
$2m:T_2-2mg=2m{a}_2\dots \dots \left(4\right)$
$m:T_1-mg=m{a}_1\dots \dots \left(5\right)$

Putting the values of $a_1$, $a_2$ and $a_3$ into eq (1) from (2), (3), (4) and (5) we get

$2[\frac{2T_2}{m}-g]=[g-\frac{T_2}{3m}]-[\frac{T_2}{2m}-g]$
$\Rightarrow \frac{4T_2}{m}+\frac{T_2}{3m}+\frac{T_2}{2m}=4g\Rightarrow T_2=\frac{24mg}{29}$

Thus, the tension acting on the mass $2m$ is $\frac{24mg}{29}N$